Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $a = \dfrac{x^2 - 6x - 7}{x^2 - 10x} \div \dfrac{6x + 6}{x - 10} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $a = \dfrac{x^2 - 6x - 7}{x^2 - 10x} \times \dfrac{x - 10}{6x + 6} $ First factor the quadratic. $a = \dfrac{(x + 1)(x - 7)}{x^2 - 10x} \times \dfrac{x - 10}{6x + 6} $ Then factor out any other terms. $a = \dfrac{(x + 1)(x - 7)}{x(x - 10)} \times \dfrac{x - 10}{6(x + 1)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac{ (x + 1)(x - 7) \times (x - 10) } { x(x - 10) \times 6(x + 1) } $ $a = \dfrac{ (x + 1)(x - 7)(x - 10)}{ 6x(x - 10)(x + 1)} $ Notice that $(x - 10)$ and $(x + 1)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac{ \cancel{(x + 1)}(x - 7)(x - 10)}{ 6x(x - 10)\cancel{(x + 1)}} $ We are dividing by $x + 1$ , so $x + 1 \neq 0$ Therefore, $x \neq -1$ $a = \dfrac{ \cancel{(x + 1)}(x - 7)\cancel{(x - 10)}}{ 6x\cancel{(x - 10)}\cancel{(x + 1)}} $ We are dividing by $x - 10$ , so $x - 10 \neq 0$ Therefore, $x \neq 10$ $a = \dfrac{x - 7}{6x} ; \space x \neq -1 ; \space x \neq 10 $